1. Solutions & Percent Concentration
Solutions, Percent Concentrations, and Dilutions
Solutions, Percent Concentrations, and Dilutions
Pima Medical Institute Online

2. Solutions Stored as a powder Injected as a solution
Some injectable drugs are stored as powders and reconstituted into a solution just before use.

3. Solutions solute solute
solvent
Substance to be dissolved
Solutions are made up of a solvent, which is the dissolving substance, and a solute, which is the dissolved substance. Often, sterile water or saline is used as solvents for the powdered drug, which is the solute. The amount of solute dissolved in a solvent is referred to as the concentration of the solution.
Sterile water or saline is often used as a solvent
Amount of solute = concentration of solution
Dissolving substance

4. Percent Concentration
1 mL
1 g
1 cc
For most solutions, keep in mind that a milliliter, a cubic centimeter, and a gram all have about the same weight. So a solution with a 1% concentration, means that there is 1 gram of the active ingredient dissolved in 100 mL of the solution.
A 15.5% concentration means that there is 15.5 grams of active ingredient dissolved in 100 mL of the solution.
100 mL
100 mL
15.5% =
1% =
1 gram
in every
in every
15.5 g

5. Self-Test: Percent Concentration

6. Practical Application
Use the percent concentration as the concentration of the drug
An 8 kg terrier usually requires 15 mg/kg of thiopental for anesthetic induction. This is an injection given to anesthetize an animal before using a gaseous anesthetic.
How much of the 2.5% drug will you draw up in a syringe?
(Body weight)(Desired volume)
Concentration of drug
Dose
2.5 x 1,000 = 2,500
( ) mg kg
When determining a patient’s dose, use the percent concentration as the concentration of the drug.
Here’s an example. An 8 kg terrier usually requires 15 mg/kg of thiopental for anesthetic induction. This is an injection given to anesthetize an animal before using a gaseous anesthetic. How much of the 2.5% drug will you draw up in a syringe?
Let’s recall our formula for dose. Dose = (body weight x desired volume)/concentration of the drug.
What is the body weight? 8 kg
What is the desired volume? 15 mg/kg
What is the concentration of the drug? 2.5%
Looking at the units, we see that the kilograms will cancel out but the milligrams and grams will not. Since we know that there are 1,000 mg in 1 gram, we can multiply 2.5 g x 1,000 which gives us 2,500 mg per 100 mL. Now, our milligram units will cancel out, leaving us with the answer unit milliliters.
Multiply then simplify the fraction. Our answer is 4.8 mL must be drawn up in a syringe for the 8 kg terrier.
(8 kg)
2,500 mg
2.5 g
100 mL mL 25
(8)(15)
4.8 mL

7. Self-Test: Percent Concentration and Dose

8. Dilute a solution to yield a weaker concentration
Dilutions
Dilute a solution to yield a weaker concentration
What you have
What you need
Weaker concentration
Sometimes, you may need a solution that you don’t have on hand. You can dilute a solution if you need a weaker concentration.
For example, if you have a 15% solution, you can create a 1% solution by adding a diluent such as water.
15 g in 100 mL
15 g in 100 mL
Diluent
1 g in 100 mL
1 g in 100 mL

9. Dilutions V1 = x 100% C1 4L V2 70% C2 What you have What you need 100%
The volume and concentration you have
The volume and concentration you need
What you have
What you need
When making weaker solutions (dilutions) from stock solutions, you may use the formula:
V1 x C1 = V2 x C2
where V1 is the volume of the stock solution used, C1 is the original concentration of the stock solution, V2 is the volume of the final concentration, and C2 is the desired concentration.
You have a 100% stock solution <<indicate C1>> and you need to prepare 4L <<indicate C2>> of a 70% solution <<indicate V2>>.
Using the dilutions formula, you can fill in the known information—C1, V2, and C2. This gives you V1 x 100% = 4L x 70%
100%
70%

10. Dilutions V1 = x 100% 4L 70% = = = = V1 x 100% 4L x 70% 100% 100%C1 V2 C2
What amount do you use? V1 = x
100% 4L
70%
V1 x 100%
4L x 70% =
100%
100%
V1 x 100
4L x 70 =
100
100
280L V1 =
Now you just need to solve for V1. V1 is the volume (or liquid amount) you need to use in order to yield 4L of a 70% solution.
Bring 100% to the other side of the equation by dividing both sides by 100%
V1 x 100% = 4L x 70%
100% %
On the left side of the equation, a percent sign appears in the numerator and the denominator. Cancel the percents. The same applies to the right side of the equation. Cancel the percents. Your answer will be in liters.
On the left side of the equation, the 100 on the top cancels out the 100 on the bottom.
V1 x 100 = 4L x 70
On the right side of the equation, 4 times 70 is 280.
V1 = 280L
100
On the right side of the equation, 280 divided by 100 is 2.8.
V1 = 2.80 L
The answer is 2.8L. You would need 2.8L of the 100% stock solution in order to make 4L of the 70% solution.
V1 = 2.8 L
100
You need 2.80L of 100% solution V1 =
2.80L

11. In order to yield 4L, how much diluent should you add?
Dilutions
2.8L + ?L
= 4L
In order to yield 4L, how much diluent should you add?
2.8L + ?L – 2.8L = 4L – 2.8L
?L = 1.2L
What you have
What you need
So, in order to yield 4L, how much of the diluent (water, in this case) should you add?
4L (70% solution) = 2.8L (100% solution) + ? (diluent)
To find our answer, subtract 2.8L from both sides.
4L – 2.8L = 2.8L + diluent – 2.8L
You have 2.8L but you need 4L.
4L – 2.8 = 1.2L
4L of 70% solution
2.8L of 100% solution
1.2L of Diluent
?L of Diluent

12. Self-Test: Percent Concentration and Dose
Let’s look at one more example.
Question: How would you prepare 100 ml of a 5% dextrose from a 50% dextrose solution?
First, use the dilution formula to identify your known information.
V1 x C1 = V2 x C2
You need 100mL of a 5% dextrose solution. Your original concentration is 50%. The only unknown left is the original volume.
V1 x 50% = 100mL x 5%
Once again, bring 50% to the right side of the equation by dividing both sides by 50%.
50% %
On the left side of the equation, a percent sign appears in the numerator and the denominator. Cancel the percents. The same applies to the right side of the equation. Cancel the percents. Your answer will be in milliliters.
On the left side of the equation, the 50 on the top cancels out the 50 on the bottom.
V1 x 50 = 100mL x 5
On the right side of the equation, 100 times 5 is 500.
V1 = 500 50
500 divided by 5 is 10. This leaves you with 10mL. You need 10mL of 50% dextrose solution in order to create 100mL of 5% dextrose solution.
V1 = 10mL
Now, how much diluent do you add to the 5mL in order to yield 100mL?
100mL – 10mL = 90mL
It can also be written this way. This formula demonstrates that you would take 10 ml of the 50% dextrose solution and add 90mL of diluent to prepare 100mL of 5 % solution.
100mL (5% solution) = 90mL (diluent) + 10mL (of 50% solution)

13. Presented by PMI Online
Resources:
jasminechemical.com
Presented by PMI Online